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# M06-16

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Math Expert
Joined: 02 Sep 2009
Posts: 55274

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16 Sep 2014, 00:27
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Difficulty:

85% (hard)

Question Stats:

45% (01:33) correct 55% (01:57) wrong based on 110 sessions

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If $$\frac{t}{u} = \frac{x}{y}$$ and $$\frac{t}{y} = \frac{u}{x}$$ and $$t$$, $$u$$, $$x$$, and $$y$$ are non-zero integers, which of the following is true?

A. $$\frac{t}{u}=1$$
B. $$\frac{y}{x}=-1$$
C. $$t = -u$$
D. $$t = \pm u$$
E. None of the above

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16 Sep 2014, 00:27
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Official Solution:

If $$\frac{t}{u} = \frac{x}{y}$$ and $$\frac{t}{y} = \frac{u}{x}$$ and $$t$$, $$u$$, $$x$$, and $$y$$ are non-zero integers, which of the following is true?

A. $$\frac{t}{u}=1$$
B. $$\frac{y}{x}=-1$$
C. $$t = -u$$
D. $$t = \pm u$$
E. None of the above

Given that: $$\frac{t}{u} = \frac{x}{y}$$ and $$\frac{t}{y} = \frac{u}{x}$$.

So, $$\frac{t}{u} = \frac{x}{y}$$ and $$\frac{t}{u} = \frac{y}{x}$$ (from 2), which means that $$\frac{t}{u}$$ and $$\frac{x}{y}$$ equal to their reciprocals: $$\frac{t}{u}=\frac{u}{t}$$ and $$\frac{x}{y}=\frac{y}{x}$$. So, $$t^2=u^2$$ and $$t^2=u^2$$, which gives $$|t|=|u|$$ (or which is the same $$t = \pm u$$) and $$|x|=|y|$$ (or which is the same $$x = \pm y$$).

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13 Feb 2015, 22:18
1
Eqn 1....Let t/u = x/y = k

Therefore t = uk, x = yk.....

Substitute in eqn 2...t/y = u/x

uk/y = u/yk

k^2 = 1

k = +/-1

so t = +/- u
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17 Feb 2015, 11:55
Bunuel In questions where they ask which is true.. are they asking for must be true or which can be true?

Asking since i solved this question as which can be true and stopped after choosing A as A was coming as true for a case where all numbers were equal.
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17 Feb 2015, 11:57
1
qw1981 wrote:
Bunuel In questions where they ask which is true.. are they asking for must be true or which can be true?

Asking since i solved this question as which can be true and stopped after choosing A as A was coming as true for a case where all numbers were equal.

Which is true = which must be true.
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10 Jun 2015, 11:54
Why are we cross-multiplying ? The question does not state that integers t,x,u,v are positive...
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10 Jun 2015, 11:56
Randude wrote:
Why are we cross-multiplying ? The question does not state that integers t,x,u,v are positive...

It does not matter for equations, it does only for inequalities.
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11 Dec 2015, 05:31
Hi,

I did the following:

From (1), ty=xu (cross multiplying)
From (2), tx=uy

so adding each side of the equalities, t(x+y)=u(x+y) so t=u (and not t= +-u)

What am I doing wrong?
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Joined: 16 Feb 2016
Posts: 49
Concentration: Other, Other

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26 Apr 2016, 18:39
m9338 wrote:
Hi,

I did the following:

From (1), ty=xu (cross multiplying)
From (2), tx=uy

so adding each side of the equalities, t(x+y)=u(x+y) so t=u (and not t= +-u)

What am I doing wrong?

t=u
would be true for negative and positive values of u.
But I can see that some people would choose E.
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Joined: 20 Jul 2012
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15 Jun 2016, 02:34
2
m9338 wrote:
Hi,

I did the following:

From (1), ty=xu (cross multiplying)
From (2), tx=uy

so adding each side of the equalities, t(x+y)=u(x+y) so t=u (and not t= +-u)

What am I doing wrong?

One explanation that i can offer here is as follows -

The equation t(x+y)=u(x+y) holds good in almost all case, however (given x,y are non zero integers), consider a scenario where x= -y.
In this case you cannot say that t=u as you cant divide both sides with (x+y) !!! why? coz its 0

Incidentally this is one of the wrong answer choices as well
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18 Feb 2017, 07:53
I still can't understand why option D is correct, even though the question is MUST BE TRUE.
Here is my thinking:

Let's say t, u, x and y are all equal to 1 (non-zero integer). Then according to option D: 1 = +-1, which is only true for positive sign, but not negative. I mean the option D would be a MUST BE TRUE answer only if it was written like this:
t = u OR t = -u.
As far as I understand t = +-u means 't = u AND t = -u'
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10 Jul 2018, 19:42
1
why is |t| = |u| the same thing as t = ±u? I don't understand that. Can someone please explain?
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Posts: 55274

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10 Jul 2018, 20:27
gdhume wrote:
why is |t| = |u| the same thing as t = ±u? I don't understand that. Can someone please explain?

|t| = |u| means that the distance from t to 0 is the same as the distance from u from 0. So, either t = s, or t = -s.
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13 Aug 2018, 07:24
m9338 wrote:
Hi,

I did the following:

From (1), ty=xu (cross multiplying)
From (2), tx=uy

so adding each side of the equalities, t(x+y)=u(x+y) so t=u (and not t= +-u)

What am I doing wrong?

I still cannot understand why this is wrong. Could someone help me?
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Joined: 29 Jun 2017
Posts: 31

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05 Sep 2018, 06:40
So, t^2=u^2 and t^2=u^2 gives |t|=|u|

Could someone please elaborate the result derived above?
Math Expert
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Posts: 55274

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05 Sep 2018, 06:45
Megha1119 wrote:
So, t^2=u^2 and t^2=u^2 gives |t|=|u|

Could someone please elaborate the result derived above?

By taking the square root.

Remember: $$\sqrt{x^2}=|x|$$.

The point here is that square root function can not give negative result: wich means that $$\sqrt{some \ expression}\geq{0}$$.

So $$\sqrt{x^2}\geq{0}$$. But what does $$\sqrt{x^2}$$ equal to?

Let's consider following examples:
If $$x=5$$ --> $$\sqrt{x^2}=\sqrt{25}=5=x=positive$$;
If $$x=-5$$ --> $$\sqrt{x^2}=\sqrt{25}=5=-x=positive$$.

So we got that:
$$\sqrt{x^2}=x$$, if $$x\geq{0}$$;
$$\sqrt{x^2}=-x$$, if $$x<0$$.

What function does exactly the same thing? The absolute value function! That is why $$\sqrt{x^2}=|x|$$
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07 Nov 2018, 10:26
Hi Bunuel and chetan2u,

I am wondering if my approach is correct? Could you verify please?

$$\frac{T}{U}$$=$$\frac{X}{Y}$$ --> $$TY-XU=0$$

$$\frac{T}{Y}$$=$$\frac{U}{X}$$ --> $$TX-UY=0$$

Combining the 2 equations:

$$TY-XU=TX-UY$$
$$TY+UY=TX+XU$$
$$Y*(T+U)=X*(T+U)$$

Can I correctly infern that either $$|Y|=|X|$$ or $$(T+U)=0$$, therefore $$|T|=|U|$$?

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07 Nov 2018, 10:37
m9338 wrote:
Hi,

I did the following:

From (1), ty=xu (cross multiplying)
From (2), tx=uy

so adding each side of the equalities, t(x+y)=u(x+y) so t=u (and not t= +-u)

What am I doing wrong?

Hi,

You will be right if (x+y) not equals to zero. But the stem says all are non zero integers that doesnt mean all are positive. so it is still possible that x is positive and y is negative with equal magnitude (or vice versa). Hence we can't just cancel out the x+y . Hope it helps.

Kudos=Thanks
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01 Dec 2018, 04:49
Bunuel,

Hi..I have a doubt in this problem. t/u=x/y and t/u=y/x which basically means that t/u is equal to x/y as well as its reciprocal (y/x) which only means that x and y has to be of the same value making t/u =1 (Option A).

2ndly, y/x=-1 which also makes x/y = -1 therby t/u=x/y and also its reciprocal (y/x) thus option B also seems plausible.

I understood your explanation for Option D but Id like to understand the gap in my reasoning if I were to choose either A or B. Thankyou
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Joined: 02 Sep 2009
Posts: 55274

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01 Dec 2018, 07:01
yashna36 wrote:
Bunuel,

Hi..I have a doubt in this problem. t/u=x/y and t/u=y/x which basically means that t/u is equal to x/y as well as its reciprocal (y/x) which only means that x and y has to be of the same value making t/u =1 (Option A).

2ndly, y/x=-1 which also makes x/y = -1 therby t/u=x/y and also its reciprocal (y/x) thus option B also seems plausible.

I understood your explanation for Option D but Id like to understand the gap in my reasoning if I were to choose either A or B. Thankyou

x/y = y/x does not mean that x = y. Cross-multiply: x^2 = y^2 --> |x| = |y|. Which means that x and y are same distance apart from 0: x = y or x = -y.
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# M06-16

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